$\pi(x)\leq \frac x{f(x)}$ for some unbounded function $f(x)$

Question

Let $\pi(x)$ denote the number of primes $\le x$. Can one prove $$\pi(x)\leq \frac x{f(x)}$$ for some function $f(x)(x\gt0)$, and $f(x)$ is unbounded?

Please do not refer to prime number theorem

From 1848 and 1850, Chebyshev proved that

$$\pi(x)\leq 1.1056\frac x{\log x}$$

Is there a possibile that the idea to apply elementary method to seek other unbounded function $f(x)$ different from $\log x$ such that $\pi(x)\leq \frac x{f(x)}$ ?

Thanks a lot!

Answer 1

You can take $f(x) = \frac{x}{\pi(x)}$ and then prove that this function is unbounded. For that it is sufficient to prove that $\lim_{x \to \infty} \frac{\pi(x)}{x} = 0$. This question has a great answer here:

prove that $\lim_{x\to\infty} \pi(x)/x=0$

Answer 2

Given $x$, let $g(x)$ denote the number of integers less than $x$ that are not divisible by any prime up to $\sqrt{\log x}$, say. Note that $\pi(x)\le g(x)$ always. It's possible, using inclusion/exclusion (the sieve of Eratosthenes), to show that $g(x) < x/f(x)$ where $f(x) = \delta\log\log x$ for some constant $\delta>0$. Look at the beginning of Chapter 3 of Montgomery/Vaughan's Multiplicative Number Theory. I. Note however that this approaches actually uses Chebyshev's theorem and its consequences (due to Mertens) as input.